A090357 Satisfies A^5 = BINOMIAL(A)^4; also equals A090356^4.

1, 4, 26, 244, 3131, 52600, 1111940, 28559320, 865622825, 30250881420, 1196941704454, 52860066623036, 2576115583371739, 137274420821505776, 7937914900025008984, 494941882189888642832, 33096552232229291234923

Offset

0,2

Comments

See comments in A090356.

Links

Table of n, a(n) for n=0..16.

Formula

G.f.: A(x)^5 = A(x/(1-x))^4/(1-x)^4.

From Peter Bala, May 26 2015: (Start)

O.g.f.: A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = Sum_{k = 1..n} k!*Stirling2(n,k)*4^k = A094417(n) = 4*A050353(n) for n >= 1.

BINOMIAL(A(x)) = exp( Sum_{n >= 1} c(n)*x^n/n ) where c(n) = (-1)^n*Sum_{k = 1..n} k!*Stirling2(n,k)*(-5)^k = A201365(n) = 5*A050353(n) for n >= 1.

A(x) = B(x)^4 and BINOMIAL(A(x)) = B(x)^5 where B(x) = 1 + x + 5*x^2 + 45*x^3 + 495*x^4 + ... is the o.g.f. for A090356. See also A019538. (End)

Mathematica

nmax = 16; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^5 - A[x/(1 - x)]^4/(1 - x)^4 + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 02 2019 *)

Prog

(PARI) {a(n)=local(A); if(n<1, 0, A=1+x+x*O(x^n); for(k=1, n, B=subst(A, x, x/(1-x))/(1-x)+x*O(x^n); A=A-A^5+B^4); polcoeff(A, n, x))}

Crossrefs

Cf. A090356; A019538, A050353, A201365.

Sequence in context: A210917 A210918 A052880 * A322766 A160886 A192546

Adjacent sequences: A090354 A090355 A090356 * A090358 A090359 A090360

Keyword

nonn,easy

Author

Paul D. Hanna, Nov 26 2003

Status

approved